Step 1

We are given the force $F = 600\;{\rm{N}}$.

We are asked the magnitude of the projection of force $F = 600\;{\rm{N}}$ along u axis.

 
Step 2

The following is the diagram of the system:

We have the vector representation of point O is $O = \left( {0i + 0j + 0k} \right){\rm{ m}}$.

We have the vector representation of point A is $A = \left( { - 2i + 4j + 4k} \right){\rm{ m}}$.

 
Step 3

The magnitude of position vector of OA is,

\[{R_{OA}} = A - O\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{l} {R_{OA}} = \left( { - 2i + 4j + 4k} \right){\rm{ m}} - \left( {0i + 0j + 0k} \right){\rm{ m}}\\ {R_{OA}} = \left( { - 2i + 4j + 4k} \right){\rm{ m}} \end{array}\]
 
Step 5

The unit vector with position vector of OA is,

\[{U_{OA}} = \frac{{{R_{OA}}}}{{\left| {{R_{OA}}} \right|}}\]
 
Step 6

Substitute the values in the above expression.

\[\begin{array}{l} {U_{OA}} = \frac{{\left( { - 2i + 4j + 4k} \right){\rm{ m}}}}{{\left[ {\sqrt {{{\left( { - 2} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 4 \right)}^2}} } \right]{\rm{ m}}}}\\ {U_{OA}} = \frac{{\left( { - 2i + 4j + 4k} \right)}}{{\sqrt {36} }}\\ {U_{OA}} = \frac{{\left( { - 2i + 4j + 4k} \right)}}{6} \end{array}\]
 
Step 7

The magnitude of the force vector along OA is,

\[{F_{OA}} = F \cdot {U_{OA}}\]
 
Step 8

Substitute the values in the above expression.

\[\begin{array}{l} {F_{OA}} = 600\;{\rm{N}} \cdot \frac{{\left( { - 2i + 4j + 4k} \right)}}{6}\\ {F_{OA}} = \left( { - 200i + 400j + 400k} \right)\;{\rm{N}} \end{array}\]
 
Step 9

The unit vector along the u axis is,

\[\begin{array}{l} {U_u} = \cos \left( {90^\circ - 30^\circ } \right)i + \left( {\cos 30^\circ } \right)j\\ {U_u} = \left( {\sin 30^\circ } \right)i + \left( {\cos 30^\circ } \right)j \end{array}\]
 
Step 10

The magnitude of the projection of force $F = 600\;{\rm{N}}$ along u axis is,

\[{F_u} = {F_{OA}} \cdot {U_u}\] Step 11

Substitute the values in the above expression.

\[\begin{array}{l} {F_u} = {F_{OA}} \cdot {U_u}\\ {F_u} = \left[ {\left( { - 200i + 400j + 400k} \right)\;{\rm{N}}} \right] \cdot \left[ {\left( {\sin 30^\circ } \right)i + \left( {\cos 30^\circ } \right)j} \right]\\ {F_u} = \left[ {\left( { - 200} \right)\left( {\sin 30^\circ } \right) + \left( {400} \right)\left( {\cos 30^\circ } \right)} \right]\;{\rm{N}}\\ {F_u} = 246.4\;{\rm{N}} \end{array}\]